Compound interest objective type question set 2

Answer: b. 12 years

Explanation:
In Compound Interest, if a sum becomes \( k \) times in \( T \) years, then it becomes \( k^n \) times in \( n \times T \) years.

Step 1: Identify the given values
Growth factor (\( k \)) = 3 (The sum triples)
Time (\( T \)) = 4 years

Step 2: Apply the Power Rule
We want the sum to become 27 times itself.
Since \( 27 = 3^3 \), we have \( n = 3 \).

Step 3: Calculate the total time
Total Time = \( n \times T \)
Total Time = \( 3 \times 4 = 12 \) years.

The sum will become 27 times in 12 years.

Answer: d. 5%

Explanation (Shortcut Method):
In Compound Interest, the difference between the interest of the first year and the second year is the interest calculated on the first year’s interest.

Step 1: Find the interest difference
Interest for 1st Year (SI) = ₹600
Interest for 2nd Year (CI) = ₹630
Difference = \( 630 – 600 = 30 \)

Step 2: Calculate the Rate (\( r \))
The difference of ₹30 is the interest earned on the first year’s interest (₹600).
\[ \text{Rate} = \left( \frac{\text{Difference}}{\text{First Year Interest}} \times 100 \right)\% \]
\[ r = \left( \frac{30}{600} \times 100 \right)\% \]
\[ r = \frac{30}{6} = 5\% \]

The rate of interest is 5% per annum.

Answer: c. Rs. 29,282

Explanation:
Given: Principal (\( P \)) = 22,000, Rate (\( r \)) = 10%, and Time (\( n \)) = 3 years.
Using the Compound Interest formula: \[ A = P \left( 1 + \frac{r}{100} \right)^n \]

Step 1: Substitute the values
\[ A = 22000 \left( 1 + \frac{10}{100} \right)^3 \]
\[ A = 22000 \left( \frac{11}{10} \right)^3 \]

Step 2: Calculate the cube
Since \( 11^3 = 1331 \) and \( 10^3 = 1000 \):
\[ A = 22000 \times \frac{1331}{1000} \]

Step 3: Solve for the final amount
\[ A = 22 \times 1331 \]
\[ A = 29282 \]
The total amount after 3 years is ₹29,282.

Answer: d. 21 years

Explanation:
In Compound Interest, if a sum becomes \( k^1 \) times in \( T \) years, it will become \( k^n \) times in \( n \times T \) years.

Step 1: Identify the Growth Factor
The sum becomes 2 times (doubles) in 3 years.
Here, \( k = 2 \) and \( T = 3 \).

Step 2: Express the Target Growth as a Power
We want the sum to become 128 times itself.
Since \( 128 = 2^7 \), our exponent \( n = 7 \).

Step 3: Calculate the Total Time
\[ \text{Total Time} = n \times T \]
\[ \text{Total Time} = 7 \times 3 = 21 \text{ years} \]

The sum will become 128 times in 21 years.

Answer: d. 21 years

Explanation:
When interest is compounded half-yearly:
1. The Rate is halved: \( R_{new} = \frac{20\%}{2} = 10\% \)
2. The Time is doubled: \( n_{new} = 2 \times 2 = 4 \text{ periods} \)

Step 1: Apply the Half-Yearly Formula
Using: \[ A = P \left( 1 + \frac{R/2}{100} \right)^{2n} \]
\[ A = 20000 \left( 1 + \frac{10}{100} \right)^4 \]

Step 2: Simplify and Calculate
\[ A = 20000 \times \left( \frac{11}{10} \right)^4 \]
\[ A = 20000 \times \frac{11 \times 11 \times 11 \times 11}{10 \times 10 \times 10 \times 10} \]
\[ A = 2 \times 14641 \]

Step 3: Final Amount
\[ A = 29282 \]
The total amount after 2 years is ₹29,282.

Answer: b. 2 year

Explanation:
Given: Principal (\( P \)) = 30,000 and Compound Interest = 3,708.
Total Amount (\( A \)) = \( 30,000 + 3,708 = 33,708 \).
Rate (\( r \)) = 6% per annum.

Step 1: Set up the Compound Interest Formula
\[ A = P \left( 1 + \frac{r}{100} \right)^n \]
\[ 33,708 = 30,000 \left( 1 + \frac{6}{100} \right)^n \]
\[ \frac{33,708}{30,000} = \left( \frac{106}{100} \right)^n \]

Step 2: Simplify the Ratio
Divide both the numerator and denominator by 3:
\[ \frac{11,236}{10,000} = \left( \frac{106}{100} \right)^n \]

Step 3: Compare Bases and Solve for \( n \)
Since \( 100^2 = 10,000 \) and \( 106^2 = 11,236 \):
\[ \left( \frac{106}{100} \right)^2 = \left( \frac{106}{100} \right)^n \]

By comparing the exponents: \( n = 2 \) years.

Answer: b. Rs. 14,405

Explanation:
When a debt is paid in equal annual installments under Compound Interest, we use the Present Value formula:
\[ P = \frac{x}{(1 + \frac{r}{100})^1} + \frac{x}{(1 + \frac{r}{100})^2} \]
Where \( P \) is the Principal (₹25,000) and \( x \) is the equal installment.

Step 1: Simplify the terms
Rate (\( r \)) = 10%. So, the factor \( (1 + \frac{10}{100}) = \frac{11}{10} \).
\[ 25000 = \frac{x}{11/10} + \frac{x}{(11/10)^2} \]
\[ 25000 = \frac{10x}{11} + \frac{100x}{121} \]

Step 2: Find a common denominator (121)
\[ 25000 = \frac{110x + 100x}{121} \]
\[ 25000 = \frac{210x}{121} \]

Step 3: Solve for \( x \)
\[ x = \frac{25000 \times 121}{210} \]
\[ x = \frac{2500 \times 121}{21} \approx 14404.76 \]
Each equal installment would be approximately ₹14,405.

Answer: b. 8 years

Explanation:
In Compound Interest, if a sum becomes \( k^1 \) times in \( T \) years, it will become \( k^n \) times in \( n \times T \) years.

Step 1: Identify the given values
The sum becomes 3 times (\( k = 3 \)) in \( T = 4 \) years.

Step 2: Express the target growth as a power
We want to find the time for the sum to become 9 times itself.
Since \( 9 = 3^2 \), our exponent (\( n \)) is 2.

Step 3: Calculate the total time
\[ \text{Total Time} = n \times T \]
\[ \text{Total Time} = 2 \times 4 = 8 \text{ years} \]

The sum of money will become 9 times the original amount in 8 years.

Answer: b. Rs. 10,648

Explanation:
Given: Principal (\( P \)) = 8,000, Rate (\( r \)) = 10%, and Time (\( n \)) = 3 years.
Using the Compound Interest formula: \[ A = P \left( 1 + \frac{r}{100} \right)^n \]

Step 1: Substitute the values
\[ A = 8000 \left( 1 + \frac{10}{100} \right)^3 \]
\[ A = 8000 \left( \frac{11}{10} \right)^3 \]

Step 2: Simplify the calculation
Since \( 11^3 = 1331 \) and \( 10^3 = 1000 \):
\[ A = 8000 \times \frac{1331}{1000} \]
Cancel the three zeros in 8,000 with the 1,000 in the denominator:
\[ A = 8 \times 1331 \]

Step 3: Final Result
\[ A = 10648 \]
The total amount after 3 years is ₹10,648.

Answer: c. Rs. 1584

Explanation:
Given: Principal (\( P \)) = 1,440 and Rate (\( r \)) = 10% for 1 year.
Using the formula: \[ A = P \left( 1 + \frac{r}{100} \right)^n \]

Step 1: Substitute the values
\[ A = 1440 \left( 1 + \frac{10}{100} \right)^1 \]
\[ A = 1440 \left( \frac{11}{10} \right) \]

Step 2: Solve for the final amount
Cancel the zero in the denominator with the zero in 1440:
\[ A = 144 \times 11 \]
\[ A = 1584 \]

The total amount after 1 year is ₹1,584.

Answer: c. Rs. 1584

Explanation:
To find the rate of interest, we compare the interest earned to the original principal.

Step 1: Find the interest amount
Difference (Interest) = \( 1498 – 1400 = 98 \)

Step 2: Set up the percentage equation
We need to find what percent of 1,400 is 98:
\[ 1400 \times \frac{r}{100} = 98 \]

Step 3: Solve for \( r \)
\[ 14r = 98 \]
Divide 98 by 14:
\[ r = \frac{98}{14} = 7 \]

The rate of interest is 7% per annum.

Answer: b. 2 year

Explanation:
Given:
Principal (\( P \)) = ₹600
Amount (\( A \)) = ₹661.50
Rate (\( R \)) = 5% per annum

Step 1: Use the Compound Interest Formula
\[ A = P \left( 1 + \frac{R}{100} \right)^n \]
\[ 661.5 = 600 \left( 1 + \frac{5}{100} \right)^n \]
\[ \frac{661.5}{600} = \left( \frac{105}{100} \right)^n \]

Step 2: Simplify the Ratio
Remove the decimal by multiplying both top and bottom by 10:
\[ \frac{6615}{6000} = \left( \frac{21}{20} \right)^n \]
Now, divide both numerator and denominator by 15:
\[ \frac{441}{400} = \left( \frac{21}{20} \right)^n \]

Step 3: Compare Bases
We know that \( 441 = 21^2 \) and \( 400 = 20^2 \):
\[ \left( \frac{21}{20} \right)^2 = \left( \frac{21}{20} \right)^n \]

Final Answer:
By comparing the exponents: \( n = 2 \) years.

Answer: d. Rs. 25,000

Explanation:
When the time period is 2 years, the difference between Compound Interest (CI) and Simple Interest (SI) can be calculated using a shortcut formula.

Formula:
\[ \text{Difference (D)} = P \left( \frac{R}{100} \right)^2 \]

Step 1: Substitute the given values
Given: Difference (\( D \)) = ₹250, Rate (\( R \)) = 10%, Time (\( n \)) = 2 years.
\[ 250 = P \left( \frac{10}{100} \right)^2 \]
\[ 250 = P \left( \frac{1}{10} \right)^2 \]

Step 2: Solve for Principal (\( P \))
\[ 250 = P \times \frac{1}{100} \]
\[ P = 250 \times 100 \]
\[ P = 25000 \]

The principal amount is ₹25,000.

Answer: b. Rs. 20,000

Explanation:
When the time period is 3 years, we use the specific difference formula to find the Principal.

Formula:
\[ \text{Difference (D)} = P \left( \frac{R}{100} \right)^2 \times \left( \frac{300 + R}{100} \right) \]

Step 1: Substitute the given values
Given: Difference (\( D \)) = ₹620, Rate (\( R \)) = 10%, Time (\( n \)) = 3 years.
\[ 620 = P \left( \frac{10}{100} \right)^2 \times \left( \frac{300 + 10}{100} \right) \]
\[ 620 = P \left( \frac{1}{100} \right) \times \left( \frac{310}{100} \right) \]

Step 2: Simplify the equation
\[ 620 = P \times \frac{310}{10000} \]
\[ P = \frac{620 \times 10000}{310} \]

Step 3: Solve for Principal (\( P \))
Since \( 620 = 2 \times 310 \):
\[ P = 2 \times 10000 \]
\[ P = 20000 \]

The principal amount is ₹20,000.

Answer: a. 2

Explanation:
According to the question, we equate the ratio of the Amount and Principal to the interest factor:

Step 1: Set up the ratio
\[ \frac{441}{400} = \left( 1 + \frac{5}{100} \right)^n \]

Step 2: Simplify the interest factor
Reducing \( 1 + \frac{5}{100} \) gives us \( \frac{105}{100} \), which simplifies to \( \frac{21}{20} \):
\[ \frac{441}{400} = \left( \frac{21}{20} \right)^n \]

Step 3: Express as a power
Recognizing that \( 441 = 21^2 \) and \( 400 = 20^2 \):
\[ \left( \frac{21}{20} \right)^2 = \left( \frac{21}{20} \right)^n \]

Final Answer:
By comparing the exponents, we get: \( n = 2 \) years.

Answer: b. 4

Explanation:
The amount becomes its 2 times in 2 years
The amount will became its 4 times which is 8000 in 2*2 = 4 years.

Answer: b. 615

Explanation:
This problem involves two steps: first finding the Principal using Simple Interest, then calculating the Compound Interest for the same Principal.

Step 1: Find the Principal (\( P \))
Given Simple Interest (\( SI \)) = ₹600, Rate = 5% (implied from \( 21/20 \) factor), and Time = 2 years.
Using: \[ SI = \frac{P \times R \times T}{100} \]
\[ 600 = \frac{P \times 5 \times 2}{100} \]
\[ 600 = \frac{P}{10} \implies P = 6000 \]

Step 2: Calculate the Compound Interest Amount (\( A \))
Using the Principal ₹6,000 at 5% for 2 years:
\[ A = 6000 \times \left( \frac{105}{100} \right)^2 \]
\[ A = 6000 \times \frac{21}{20} \times \frac{21}{20} \]
\[ A = 15 \times 441 = 6615 \]

Step 3: Final Compound Interest (\( CI \))
\[ CI = A – P \]
\[ CI = 6615 – 6000 = 615 \]

The compound interest is ₹615.

Answer: b. 10

Explanation:
We need to find the rate of interest (\( r \)) where the Principal (\( P \)) is ₹2,000, the Amount (\( A \)) is ₹2,662, and the Time (\( n \)) is 3 years.

Step 1: Set up the formula
Using the formula: \[ A = P \left( 1 + \frac{r}{100} \right)^n \]
\[ 2662 = 2000 \left( 1 + \frac{r}{100} \right)^3 \]

Step 2: Simplify the ratio
Divide both sides by 2,000:
\[ \frac{2662}{2000} = \left( 1 + \frac{r}{100} \right)^3 \]
\[ \frac{1331}{1000} = \left( 1 + \frac{r}{100} \right)^3 \]

Step 3: Express as a cube
We know that \( 1331 = 11^3 \) and \( 1000 = 10^3 \):
\[ \left( \frac{11}{10} \right)^3 = \left( 1 + \frac{r}{100} \right)^3 \]

Step 4: Solve for \( r \)
Removing the cubes from both sides:
\[ \frac{11}{10} = 1 + \frac{r}{100} \]
\[ \frac{11}{10} – 1 = \frac{r}{100} \implies \frac{1}{10} = \frac{r}{100} \]
\[ r = 10 \]

The rate of interest is 10% per annum.

Answer: b. 3 years

Explanation:
Given:
Principal (\( P \)) = ₹8,000
Amount (\( A \)) = ₹9,261
Rate (\( R \)) = 5% per annum

Step 1: Use the Compound Interest Formula
\[ A = P \left( 1 + \frac{R}{100} \right)^n \]
\[ 9261 = 8000 \left( 1 + \frac{5}{100} \right)^n \]

Step 2: Simplify the Ratio
Move the Principal to the other side to isolate the time factor:
\[ \frac{9261}{8000} = \left( \frac{105}{100} \right)^n \]
\[ \frac{9261}{8000} = \left( \frac{21}{20} \right)^n \]

Step 3: Recognize the Powers
We can observe that:
\( 21^3 = 9261 \)
\( 20^3 = 8000 \)

So, \[ \left( \frac{21}{20} \right)^3 = \left( \frac{21}{20} \right)^n \]

Final Answer:
By comparing the exponents: \( n = 3 \) years.

Answer: d. 13,230

Explanation:
Given:
Principal (\( P \)) = ₹12,000
Rate (\( R \)) = 5% per annum
Time (\( n \)) = 2 years

Step 1: Use the Compound Interest Formula
\[ A = P \left( 1 + \frac{R}{100} \right)^n \]
\[ A = 12000 \left( 1 + \frac{5}{100} \right)^2 \]

Step 2: Simplify the interest factor
\[ A = 12000 \left( \frac{105}{100} \right)^2 \]
Simplifying \( \frac{105}{100} \) to \( \frac{21}{20} \):
\[ A = 12000 \times \frac{21}{20} \times \frac{21}{20} \]

Step 3: Calculate the final amount
First, cancel the zeros and divide 120 by 4:
\[ A = 30 \times 21 \times 21 \]
\[ A = 30 \times 441 \]
\[ A = 13230 \]

Final Answer:
Mahesh will get ₹13,230 at the end of two years.